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Moving Object, Stationary Lens

A convex lens...

Question

A convex lens forms a real image with magnification $m_{1}$ on a screen. Now, the screen is moved by a distance x and the object is also moved so as to obtain a real image with magnification $m_{2} (> m_{1})$ on the screen. Then, the focal length of the lens is

A

[\frac{m_{1}}{m_{2}}] x

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B

[\frac{m_{2}}{m_{1}}] x

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C

x (m_{2} - m_{1})

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D

\frac{x}{(m_{2} - m_{1})}

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Solution

The correct option is C $\frac{x}{(m_{2} - m_{1})}$
$\begin{matrix} The magnification m = \frac{v}{f} - 1 in case of displacement method. case 1 V = V m_{1} = \frac{V}{f} - 1 V = (m 1 + 1) f \end{matrix}$
$\begin{matrix} caseil & v = v + x m_{2} & = \frac{v + x}{f} - 1 v & = (m_{2} + 1) f - x \Rightarrow & (m_{1} + 1) f = (m_{2} + 1) f - x m_{1} f + f = m_{2} f + f - x f = \frac{x}{m_{2} - m_{1}} \end{matrix}$

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